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I am wondering if it is always possible to find disjoint sets on any manifold such that these sets are balls when mapped to their locally Euclidean space $such$ $that$ there are an infinite number of such sets.

The result is, for example, obvious when the manifold is itself Euclidean; not sure if this is true in general.

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What does it mean "are balls when mapped to their locally Euclidean space"? It seems to me that this condition is not well defined. –  Emanuele Paolini Jan 29 '13 at 21:12
    
Every smooth manifold has a countable basis of regular coordinate balls. A smooth coordinate ball is a smooth coordinate domain whose image under a smooth coordinate map is a ball in ordinary Euclidean space. –  Squirtle Jan 29 '13 at 21:18
    
But the same manifold can have different coordinate maps, and hence a set could be mapped to a ball by a coordinate map but not by another. –  Emanuele Paolini Jan 29 '13 at 21:23

1 Answer 1

You could always look at an open $\Bbb R^n$-homeomorphic subset of your manifold and say "this is basically Euclidian, so I can find an infinitude of balls within this subset." Then you're done.

Also, note that for any open subset of a manifold homeomorphic to Euclidian space, there is no unique such homeomorphism, so a set that looks like a ball under one homeomorphism can look like a cube under another.

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